Multiplication Theorem on Probability

Turning the Definition Around

The conditional-probability definition $P(E|F) = \dfrac{P(E \cap F)}{P(F)}$ answers "what is the conditional?" — but rearranged, it answers the more common question "what is the probability that both events occur?":

$$P(E \cap F) = P(E)\,P(F|E) = P(F)\,P(E|F)$$

(each form valid when the conditioning event has nonzero probability). This is the multiplication rule of probability. The event $E \cap F$ is often written simply $EF$.

Reading it as a story

$P(E)\,P(F|E)$ reads: probability the first thing happens, times the probability the second happens given the first already has. This is exactly how sequential experiments unfold — draw a ball, the urn changes, draw again. The rule formalises "update the odds after each step, then multiply along the chain."

Extension to three or more events

$$P(E \cap F \cap G) = P(E)\;P(F|E)\;P(G|E \cap F)$$

and similarly for longer chains — each factor conditions on everything drawn so far.

Key Point (without vs with replacement): the multiplication rule earns its keep in without-replacement problems, where each draw changes the pool: both counts (favourable and total) shrink as you go. With replacement, the pool resets and the factors are all equal — a preview of independence, next section. The single most common error is forgetting to update BOTH the numerator and the denominator after each draw.

Solved Examples

Example 1: Two blacks, no replacement

An urn contains $10$ black and $5$ white balls. Two balls are drawn one after the other without replacement. Find the probability that both are black.

Solution:

1. First draw: $P(E) = \dfrac{10}{15}$.
2. Second draw, updated urn ($9$ black among $14$): $P(F|E) = \dfrac{9}{14}$.
3. Multiply: $\dfrac{10}{15} \times \dfrac{9}{14} = \dfrac{3}{7}$.

Answer: $\dfrac{3}{7}$.

Example 2: King, king, ace

Three cards are drawn successively without replacement from a well-shuffled pack. Find the probability that the first two are kings and the third is an ace.

Solution:

1. Chain of three: $P(K) = \dfrac{4}{52}$; $P(K|K) = \dfrac{3}{51}$ (three kings left in $51$); $P(A|KK) = \dfrac{4}{50}$ (all four aces still there, $50$ cards left).
2. Multiply: $\dfrac{4}{52} \times \dfrac{3}{51} \times \dfrac{4}{50} = \dfrac{2}{5525}$.

Answer: $\dfrac{2}{5525}$ — note the third factor's numerator did NOT shrink: no aces had been removed.

Example 3: Both aces

Two cards are drawn without replacement. Find the probability that both are aces.

Solution:

1. Chain: $\dfrac{4}{52} \times \dfrac{3}{51}$.
2. Simplify: $\dfrac{1}{13} \times \dfrac{1}{17} = \dfrac{1}{221}$.

Answer: $\dfrac{1}{221}$. (With replacement the answer would be $\dfrac{1}{13} \times \dfrac{1}{13} = \dfrac{1}{169}$ — replacement always makes repeats more likely… than without? Compare: $\dfrac{1}{169} > \dfrac{1}{221}$: yes, resetting the pool keeps all four aces available.)

Example 4: All three good

A box holds $15$ oranges of which $12$ are good. Three oranges are drawn without replacement. Find the probability that all three are good.

Solution:

1. Chain: $\dfrac{12}{15} \times \dfrac{11}{14} \times \dfrac{10}{13}$.
2. Compute: $\dfrac{44}{91}$.

Answer: $\dfrac{44}{91}$ — both numbers fall by one at each step: good oranges and total oranges.

Example 5: Order matters in the chain

An urn has $5$ red and $3$ black balls. Two are drawn without replacement. Find the probability that the first is red and the second is black.

Solution:

1. Chain in the stated order: $P(R_1) = \dfrac{5}{8}$; $P(B_2|R_1) = \dfrac{3}{7}$ (all three blacks remain, seven balls left).
2. Multiply: $\dfrac{15}{56}$.

Answer: $\dfrac{15}{56}$ — the chain follows the stated sequence; "one red and one black in any order" would add the reverse chain $\dfrac{3}{8}\times\dfrac{5}{7}$ as well.

Example 6: None defective

Of $10$ bulbs, $3$ are defective. Three bulbs are drawn without replacement. Find the probability that none is defective.

Solution:

1. Chain through the good bulbs: $\dfrac{7}{10} \times \dfrac{6}{9} \times \dfrac{5}{8}$.
2. Compute: $\dfrac{7}{24}$.

Answer: $\dfrac{7}{24}$.